The standard reduction potentials of Cu^2+ (aq) Cu(s) and Ag^+ (aq)Ag(s) are +0.

Question

The standard reduction potentials of Cu^2+ (aq) Cu(s) and Ag^+ (aq)Ag(s) are +0.34 and +0.60 volts, respectively. Determine the potential, E_cells (in volts) for the following cell at 25.0 C: Cu(s) Cu^2+ (0.250 M) (Ag^-(0.0010 M) Ag(s) A galvanic cell is composed of these two half cells, with the standard reduction potential shown: Cu^2+ (aq) + 2e rightwardsharpoonoverleftwardsharpoon Cu (s) +0.34 V Fe^2+ (aq) + 3e rightwardsharpoonoverleftwardsharpoon Fe(s) +0.77V what is the standard free energy change for the cell reaction of this galvanic cell? Using the reduction potentials given, calculate the equilibrium constant, K, at 25 degree C for the reaction: Ag^+ (aq) + Fe^2+ (aq) rightwardsharpoonoverleftwardsharpoon Ag(s) + Fe^Fe^2+ (aq) + e rightwardsharpoonoverleftwardsharpoon Fe^2+ (aq) +0.77V Ag^+ (aq) + e rightwardsharpoonoverleftwardsharpoon Ag(s) +0.80 V The equilibrium constant, K_c, was found to be 2.4 times 10^at 25 degree C for the reaction:2X(s) + 3Y^2+ (aq) rightwardsharpoonoverleftwardsharpoon 2X^3+ (aq) + 3Y(s) What is the standard reduction potential for this reaction? A galvanic cell is composed of these two half-cells, with the standard reduction potentials shown: Co^2 + (aq) + 2 e^- rightwardsharpoonandleftwardsharpoon Co(s) -0.28 volt Cd^2+ (aq) + 2 e^- rightwardsharpoonandleftwardsharpoon Cd(s) -0.40 volt The actual concentrations are: [Co^2+ (aq)] = 0.00100 M, [Cd^2 +] = 0.100 M. What is the potential of this galvanic cell? Using the standard reduction potentials Ni^2+ (aq) + 2 e^- rightwardsharpoonandleftwardsharpoon Ni(s) -0.25 volt Fe^3- (aq) + e^- rightwardsharpoonandleftwardsharpoon Fe^2+ (aq) +0.77 volt Calculate the value of E degree _cell for the cell with the following reaction: Ni^2+ (aq) + 2 Fe^2+ (aq) rightarrow NI(s) + 2 Fe^3+ (aq) Using the standard reduction potentials Au^3 + (aq) + 3 e^- rightwardsharpoonandleftwardsharpoon Au (s) +1.42 volt Ca^2+ (aq) + 2 e^- rightwardsharpoonandleftwardsharpoon Ca(s) -2.76 volt Calculate the value of E degree _cell for the cell reaction: 2 Au (s) + 3 Ca^2+ (aq) rightarrow 2 Au^3+ (aq) + 3 Ca(s) The electrode for which the standard reduction potential of 0.00 V is assigned making it the electrode designated as the the half reaction. a. Cd^2+ (aq) + 2e^- rightwardsharpoonoverleftwardsharpoon Cd (s) b. 2 H^+ (aq) + 2e^- rightwardsharpoonoverleftwardsharpoon H_2 (g) c. Ag^+ (aq) + e^- rightwardsharpoonoverleftwardsharpoon Ag(s) d. Au^3+ (aq) + 3e^- rightwardsharpoonoverleftwardsharpoon Au(s) e. 2 NH^+_4 (aq) + 2e^- rightwardsharpoonoverleftwardsharpoon H_2 (g) + 2 NH_3 (g) A galvanic cell consists of a Cd(s)|Cd^2+ (aq) half-cell and a Zn(s)|Zn^2+ (aq) half-cell connected by a salt bridge. Reduction The cell can be represented in standard notation as: a. Cd(s)|Cd^2+ (aq)|Zn(s)|Zn^2+ (aq) b. Zn(s)|Zn^2+ (aq)||Cd(s)|Cd^2+ (aq) c. Zn(s)|Zn^2+ (aq)||Cd^2+ (s)|Cd (aq) d. Zn^2+ (aq)|Zn(s)||Cd(s)|Cd^2+ (aq) e. Cd^2+ (aq)|Cd(s)||Zn(s)|Zn^2+ (aq) The half-reactions in nicad (nickel-cadmium) batteries are: Cd(OH)_2 (s) + 2e rightarrow Cd(s) + 2 OH^- (aq) E degree = -0.403 V 2 NiO (OH) (s) + 2 H_2O (l) + 2e rightarrow 2 Ni(OH)_2 (s) + 2 OH (aq) E degree = 1, 32 V the net ionic equation for the cell reaction and calculate the value of E degree_cell.

Explanation / Answer

1) Cu(s) + 2Ag+(aq) --------> Cu2+ (aq) + 2Ag(s)

Nernst equation is

E(cell) = E°cell - (0.0592/n)lnQ

E°Cell = E°cathode - E°anode

= 0.80V - 0.34V

= 0.46V

Number of electron transfer,n = 2

Q = [ Cu2+]/[Ag+]^2

= 0.250M / (0.0010M)^2

= 250000

Substituting the values

E(cell) = 0.46V - (0.0592V/2) log ( 250000)

= 0.46V - 0.160V

= 0.30 V

2) G° = -nFE°

n= 2

E° = 0.77V - 0.34V

= 0.43V

Faraday constant , F = 96485.3 Coulomb/mol

G° = -(2 × 96485.3(C/mol) × 0.43V)

= - 82.98 KJ

3) G° = - RTlnK

G° = -nFE°

= - 2×96485.3(C/mol)× ( 0.80V - 0.77V)

= -5.78KJ

-5.78KJ= - 8.314(J/K mol) × 298.15K ×2.303 logK

log K = 5.78KJ/5.71KJ

K = 10.28

4) K = 2.4×10^5

G° = - RTlnK

G° = - 8.314(J/mol K ) × 298.15K ×2.303log ( 2.4×10^5 )

= -30.71KJ

G° = - nFE°

E° = - G°/nF

E° = -(-30710J / (6 × 96485.3 C/mol))

   = + 30710 J/578911.8C

   =0.053V

=

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